A principal-angle bound encloses the largest selected generalized Ritz value across a spectral gap

By [F2], write a B-unit vector $x\in S$ as $x=u+e$ with $u=P_{U}x\in U$ and $e=(I-P_U)x\in U^{{\perp }_B}$. In the B-orthonormal generalized eigenbasis, if $x={\sum }_i{a}_i{u}_i$, then $x^{\ast }Bx={\sum }_i\mathrm{\mid }{a}_i{\mathrm{\mid }}^2$ and $x^{\ast }Ax={\sum }_i{\lambda }_i\mathrm{\mid }{a}_i{\mathrm{\mid }}^2$. Thus $u^{\ast }Au\le {\lambda }_q{u}^{\ast }Bu$ and $e^{\ast }Ae\le {\lambda }_n{e}^{\ast }Be$; the cross terms vanish since $u_i^{\ast }A{u}_j={\lambda }_j{u}_i^{\ast }B{u}_j=0$ for $i\le q<j$. Therefore $x^{\ast }Ax\mathrm{/}{x}^{\ast }Bx\le {\lambda }_q\mathrm{\parallel }u{\mathrm{\parallel }}_B^2+{\lambda }_n\mathrm{\parallel }e{\mathrm{\parallel }}_B^2={\lambda }_q+({\lambda }_n-{\lambda }_q)\mathrm{\parallel }e{\mathrm{\parallel }}_B^2$, which is at most the stated angle bound.

Diagonalizing the finite Hermitian compression shows that ${\theta }_q(S)$ is the maximum Rayleigh quotient on $S$, so step 1.1 gives its upper bound. The lower bounds ${\lambda }_j\le {\theta }_j(S)$ are [L1]. The strict gap isolates the first q eigenspaces, so their sum is unique; no step asserts an iteration rate.